Example of Finding Roots

Finding Roots

METHOD: To find the roots of a polynomial, first try to factor it. This which works in many examples in calculus courses. For quadratic polynomials, or quadratic factors in polynomials, use the quadratic formula.

There is no general method for finding the root of an arbitrary polynomial. The best thing to do is to find a numerical approximation of the root using a calculator or computer.

Example 1. Find the roots of

x² + 3 x + 2.

Solution: Try factoring. 2 = 1^.2 or (-1)(-2). The only possibilities are (x - 1)(x -2) and (x + 1)(x + 2). The second one works, so x² + 3 x + 2 = (x + 1)(x + 2) and the roots are -1 and -2.

Example 2. Find the roots of

x² + 4 x + 2.

Solution: Try factoring. 2 = 1^.2 or (-1)(-2). The only possibilities are (x - 1)(x -2) and (x + 1)(x + 2). None works, so use the quadratic formula: The roots are

[-4 ± sqrt( 4² - 4^.1^.2 )] / 2^.1
= ( -4 ± 2 sqrt(2) ) / 2
= -2 + sqrt(2) and -2 - sqrt(2).

Example 3. Find the roots of

3 x² + x + 6.

Solution: Factoring doesn't work (trust us!), so use the quadratic formula: The roots are

[-1 ± sqrt( 1² - 4^.3^.6 )] / 2^.3
= ( -1 ± sqrt(-71) ) / 6
= not a real number.

We conclude the polynomial has no real roots but there are two complex roots, namely x = ( -1 + sqrt(71)i ) / 6 and x = ( -1 + sqrt(71)i ) / 6. We will not be interested in complex roots in CalculusQuest^TM.

Example 4. Find the roots of

x³ - 2 x² - x - 2.

Solution: Try factoring out a linear term, or use trial and error. Say you find the factor of (x + 2). Then write x³ - 2 x² - x - 2 = (x +2)(x² - 1). The roots, then, are x = 2, x = -1, and x = 1.

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